Question

The brown ring in the detection of $N{O}_3^{-}$ ion is due to the formation of

• A. $N{O}_2$
• B. $N_2{O}_3$
• C. $N_2{O}_5$
• D. $FeS{O}_4.NO$

Answer 1

The correct option is D

$FeS{O}_4.NO$

During the brown ring test $FeS{O}_4.NO$ is formed from brown colored ion $Fe(H_{2}O{)}_5.NO{]}^{+2}$

$N{O}_3^{-}+3F{e}^{+2}+4H^{⨁}\to NO+3F{e}^{+3}+2H_{2}O$

$\left[Fe\right(H_{2}O{)}_6{]}^{+2}+NO\to \underset{NitrosoFerrousIon\left(brown\right)}{\left[Fe\right(H_{2}O{)}_5.NO{]}^{+2}}+H_{2}O$